A point charge $$+q$$ is placed at the origin. A second point charge $$+9q$$ is placed at $$(d, 0, 0)$$ in Cartesian coordinate system. The point in between them where the electric field vanishes is:

Let the first charge $$+q$$ be at the origin $$O(0,0,0)$$ and the second charge $$+9q$$ be at $$P(d,0,0)$$ on the x-axis.

Because both charges are positive, the electric field produced by each charge at any point on the line joining them points away from the respective charge.
Hence, at a point in between the two charges the field contributions will be in opposite directions and can cancel out.

Assume the required point $$R$$ is at $$x$$ (with $$0 \lt x \lt d$$).
Position of $$R$$: $$(x,0,0)$$.

Magnitude of field at $$R$$ due to $$+q$$ at the origin:
$$E_1 = \frac{kq}{x^{2}}$$ directed along $$+\hat{i}$$.

Magnitude of field at $$R$$ due to $$+9q$$ at $$P(d,0,0)$$:
$$E_2 = \frac{k(9q)}{(d-x)^{2}}$$ directed along $$-\hat{i}$$ (to the left).

For the net field to vanish, the magnitudes must be equal:
$$E_1 = E_2 \Longrightarrow \frac{kq}{x^{2}} = \frac{k(9q)}{(d-x)^{2}}$$.

Cancelling the common factors $$kq$$ gives
$$\frac{1}{x^{2}} = \frac{9}{(d-x)^{2}}$$.

Taking square roots:
$$\frac{1}{x} = \frac{3}{d-x} \Longrightarrow d - x = 3x$$.

Solving for $$x$$:
$$d = 4x \Longrightarrow x = \frac{d}{4}$$.

Therefore the electric field is zero at the point $$(d/4, 0, 0)$$.

Hence, the correct option is Option B.